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xxMikexx [17]
1 year ago
6

The internal energy of a gas is 500 J. The gas is compressed adiabatically, and its volume decreases by 100c * m ^ 3 If the pres

sure applied on the gas during compression is 3.00 atm, what is the internal energy of the gas after the adiabatic compressio
​
Physics
1 answer:
irinina [24]1 year ago
3 0

The internal energy of the gas after the adiabatic compression will be 30.398 × 10⁶ J

<h3>What is work done by gas?</h3>

When energy is moved from one store to another, work is completed. Work done on the gas is taken as -ve.

Given data;

pressure(P)=3.0 atm = 303975 N/m²

The initial volume, V₁

work is done on the gas., W=?(-ve)

Change in heat, ΔQ=0

Change in the internal energy of the gas., ΔE

The work done on the gas;

W = -PΔV

W= - 303975 N/m² × 100 cm³

W = - 30.3 × 10⁶ J

The internal energy is found as;

ΔE=q+w

ΔE= 0-30.3 × 10⁶ J

ΔE= -30.3 × 10⁶ J

E₂-E₁= -30.3 × 10⁶ J

E₂ = -30.3  × 10⁶ J +500

E₂ = 30.398 × 10⁶ J

Hence, the internal energy of the gas after the adiabatic compression will be 30.398 × 10⁶ J

To learn more about the work done by gas refer to;

brainly.com/question/12539457

#SPJ1

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Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

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Which statement is true?
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In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true fo
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Energy is released in the reaction

Explanation:

In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.

This indicates that the reaction is an exothermic or exergonic reaction.

These reaction types are accompanied by release of energy.

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3 years ago
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The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

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3 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
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Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

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(b) A 22 g song sparrow, m = 22 g = 0.022 kg

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Hence, this is the required solution.

5 0
3 years ago
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